Re: Weird trigonometric integral and Simplification question

*To*: mathgroup at smc.vnet.net*Subject*: [mg31789] Re: [mg31777] Weird trigonometric integral and Simplification question*From*: BobHanlon at aol.com*Date*: Sun, 2 Dec 2001 04:25:05 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

In a message dated 2001/12/1 4:50:18 AM, bta at attewode.com writes: >Simplification question > > If I set lambda=(1+Sqrt[5])/2, and then ask Mathematica to >Expand[lambda^7] or ask for any other power of lambda, Mathematica >gives a nice result that I can easily rewrite by hand in the form a + b >lambda where a and b are integers. Is there any way to get Mathematica >to >write the result in this form? (I'm not interested in the answer involving >the Fibonacci sequence, I'm interested in learning how to manipulate >expressions in Mathematica.) Once you give lambda a value you cannot express anything literally in terms of lambda since Mathematica will substitute the assigned value. However, the symbol GoldenRatio which is equivalent to your lambda can be used {GoldenRatio, FunctionExpand[GoldenRatio]} {GoldenRatio, (1/2)*(1 + Sqrt[5])} Clear[f]; f[n_Integer?NonNegative] := Module[{a,b, lambda = (1+Sqrt[5])/2, soln1, soln2}, Off[Solve::svars]; soln1 = FullSimplify[Solve[a+b*lambda == lambda^n, {a,b}][[1]]]; soln2 = Solve[Coefficient[FullSimplify[a /. soln1 /. Sqrt[5] -> x], x] == 0, b][[1]]; On[Solve::svars]; Simplify[a+b*GoldenRatio /. Join[soln1 /. soln2, soln2]]]; f[7] 8 + 13*GoldenRatio Table[{GoldenRatio^n,"==", t=f[n], FullSimplify[GoldenRatio^n - t]\[Equal]0}, {n, 0, 7}]//TableForm > > Similarly if I ask Mathematica for Inverse[{{a,b},{c,d}}] it gives >a >nice result, but how do I get it to factor out a d-b c? > inv = Inverse[{{a,b},{c,d}}] If you factor something out of a matrix, Mathematica will immediately carry out the multiplication or division. You can use an undefined operator to delay the operations CircleTimes[(t=Denominator[inv[[1,1]]]) * inv, 1/t] inv == Times@@% Bob Hanlon Chantilly, VA USA